The given differential equation is:

$y'' + 9y = \sin(3x)$

We are tasked with solving this using the method of variation of parameters.

Step 1: Solve the homogeneous equation

First, solve the homogeneous part of the equation:

$y'' + 9y = 0$

The characteristic equation for this is:

$r^2 + 9 = 0$

This gives:

$r = \pm 3i$

Thus, the general solution to the homogeneous equation is:

$y_h(x) = c_1 \cos(3x) + c_2 \sin(3x)$

Step 2: Particular solution using variation of parameters

The particular solution, $y_p(x)$, is found by assuming it has the form:

$y_p(x) = u_1(x) \cos(3x) + u_2(x) \sin(3x)$

where $u_1(x)$ and $u_2(x)$ are functions to be determined.

We need to compute $u_1(x)$ and $u_2(x)$ using the following formulas derived from the method of variation of parameters:

$u_1'(x) = \frac{-y_2(x) g(x)}{W(y_1, y_2)}$

$u_2'(x) = \frac{y_1(x) g(x)}{W(y_1, y_2)}$

where:

• $y_1(x) = \cos(3x)$,
• $y_2(x) = \sin(3x)$,
• $g(x) = \sin(3x)$ is the non-homogeneous term, and
• $W(y_1, y_2)$ is the Wronskian of $y_1$ and $y_2$.

Step 3: Calculate the Wronskian

The Wronskian is given by:

$W(y_1, y_2) = y_1(x) y_2'(x) - y_1'(x) y_2(x)$

We know that:

• $y_1(x) = \cos(3x)$,
• $y_1'(x) = -3 \sin(3x)$,
• $y_2(x) = \sin(3x)$,
• $y_2'(x) = 3 \cos(3x)$.

Thus,

$W(y_1, y_2) = \cos(3x) \cdot 3 \cos(3x) - (-3 \sin(3x)) \cdot \sin(3x) = 3[\cos^2(3x) + \sin^2(3x)] = 3$

Step 4: Compute $u_1'(x)$ and $u_2'(x)$

Using the formulas for $u_1'$ and $u_2'$:

$u_1'(x) = \frac{-\sin(3x) \cdot \sin(3x)}{3} = \frac{-\sin^2(3x)}{3}$

$u_2'(x) = \frac{\cos(3x) \cdot \sin(3x)}{3} = \frac{\sin(3x) \cos(3x)}{3}$

Step 5: Integrate to find $u_1(x)$ and $u_2(x)$

Now integrate $u_1'(x)$ and $u_2'(x)$:

$u_1(x) = \int \frac{-\sin^2(3x)}{3} dx$

Using the identity $\sin^2(3x) = \frac{1 - \cos(6x)}{2}$, we get:

$u_1(x) = \int \frac{-1 + \cos(6x)}{6} dx = \frac{-x}{6} + \frac{\sin(6x)}{36}$

Next,

$u_2(x) = \int \frac{\sin(3x) \cos(3x)}{3} dx$

Using the identity $\sin(3x) \cos(3x) = \frac{1}{2} \sin(6x)$, we get:

$u_2(x) = \int \frac{1}{6} \sin(6x) dx = \frac{-\cos(6x)}{36}$

Step 6: Form the particular solution

Now, substitute $u_1(x)$ and $u_2(x)$ into the expression for $y_p(x)$:

$y_p(x) = \left(\frac{-x}{6} + \frac{\sin(6x)}{36}\right) \cos(3x) + \left(\frac{-\cos(6x)}{36}\right) \sin(3x)$

Final Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:

$y(x) = y_h(x) + y_p(x)$

$y(x) = c_1 \cos(3x) + c_2 \sin(3x) + \left(\frac{-x}{6} + \frac{\sin(6x)}{36}\right) \cos(3x) + \left(\frac{-\cos(6x)}{36}\right) \sin(3x)$

This is the complete solution to the equation.

Would you like more details on any step? Here are some related questions:

1. What is the variation of parameters method used for in differential equations?
2. Can we apply the method of undetermined coefficients to this equation?
3. How do you calculate the Wronskian for different functions?
4. What other methods exist to solve non-homogeneous differential equations?
5. How do trigonometric identities help simplify integrals like $\sin^2(x)$?

Tip: Always check if a simpler method (e.g., undetermined coefficients) can solve the equation before applying variation of parameters!